Important Questions for CBSE Class 6 Maths Chapter 10 - Mensuration

Very Short Answer Questions. 1 Mark 

Q1. Length of boundary of a plane figure is called_____

Ans: 

Perimeter

Q2. Perimeter of rectangle \[=\] ____?

Ans:

We know that Perimeter of rectangle $=2(l+b),$

Here, 

\[\begin{align} & l=\text{length} \\ & b=\text{breadth} \\ \end{align}\]

Q3. Perimeter of square = ___?

Ans: 

We know that perimeter is the length of the boundaries of a figure.

Square has four sides. 

Therefore, Perimeter of square $=4\times \text{side}$

Q4. The ratio of circumference of a circle and it’s ______ is always constant.

Ans:

The ratio of circumference of a circle and it’s diameter is always constant.

We know that, circumference of a circle \[=2\pi r\]

Diameter of a circle $=d=2r$

Ratio will be \[\begin{align} & \frac{2\pi r}{2r} \\ & =2 \\ \end{align}\]

Hence, the ratio is constant.

Q5. Circumference of a circle, $C=$ ________?

Ans: 

We know that circumference is the total length of the boundary of a figure.

So, $C=2\pi r$

Q6. Value of $\pi $:

(a) In fractions is ________

(b) In decimals is ________

Ans:

(a) Value of $\pi $ in fraction is $\frac{22}{7}.$

(b) Value of $\pi $ in decimals is $3.14$

Q7. Area of rectangle \[=(a)\_\_\_\_\_\_\_\times (b)\_\_\_\_\_\_\text{square}\text{units}.\]

Ans:

We know that area of rectangle $=\text{length}\times \text{breadth}$

Therefore,

(a) Length

(b)Breadth

Q8. Area of square \[=(a)\_\_\_\_\_\_\_\times (b)\_\_\_\_\_\_\text{square}\text{units}.\]

Ans: 

We know that area of a square is $=\text{side}\times \text{side}$

Therefore,

(a) Side

(b) Side

Short Answer Questions.                                         2 Marks

Q1. Length and breadth of a rectangle are \[12.3\text{cm}\] and \[9.6\text{cm}\]respectively. Find its perimeter.

Ans:

Given: Length $=12.3\text{cm}$

Breadth $=9.6\,\,\text{cm}$

We know perimeter $=2(l+b)$

\[\begin{align} & =2(12.3+9.6) \\ & =31.5\text{cm} \\ \end{align}\]

Q2. Find the cost of fencing a square field of \[175\text{m}\] at Rs.\[60\text{ per metre}.\]

Ans:

Given: Perimeter of square field $=175\text{m}$

We need to find the cost of fencing the field.

Perimeter $=175\,\,\text{m}$

Cost of fencing per metre $=\text{Rs}\text{.}60$

So, total cost of fencing will be

\[\begin{align} & =175\times 60 \\ & =\text{Rs}\text{. 10,500} \\ \end{align}\]

Q3. Find the perimeter of a square whose has sides measuring $6.4\text{cm}$

Ans:

Given: Side of square $=6.4\text{cm}$

We need to find the perimeter of the square.

We know that the perimeter of the square $=4\times \text{side}$
So, Perimeter of square will be

\[\begin{align} & \text{P}=4\times 6.4 \\ & =25.6\text{cm} \\ \end{align}\]

Q4. Find circumference of a circle of radius \[13\text{cm}.\]

Ans:

Given: Radius $=13\text{cm}$

We need to find the circumference of the circle.

We know, Circumference $=2\pi r$

$\begin{align} & =2\times \frac{22}{7}\times 13 \\ & =81.714\text{cm} \\ \end{align}$

Q5. Find the diameter of a circle whose circumference is \[88\text{cm}.\]

Ans: 

Given: Circumference of a circle $=88\text{cm}$

We need to find the diameter of the circle.

We know that circumference $=\pi d$

\[\begin{align} & \Rightarrow d=\frac{C}{\pi } \\ & =\frac{88}{\pi } \\ & =\frac{88}{22}\times 7 \\ & =4\times 7 \\ & =28\text{cm} \\ \end{align}\]

Short Answer Questions. 3 Marks

Q1. The length and breadth of a rectangle field are \[340\text{m}\] and \[160\text{m}\text{.}\] A watchman walked $5$ rounds around the field. Find the distance covered by him.

Ans:

Given: length of rectangular field, $l=340\text{m}$

Breadth of rectangular field, $b=160\text{cm}$

We need to find the distance covered by the watchman.

We know that Perimeter, $\text{P}=2(l+b)$

\[\begin{align} & =2(340+160) \\ & =2(500) \\ & =1000\text{cm} \\ \end{align}\]

This is the length of one round but watchman walked 5 rounds.

So, distance covered by watchman will be

\[\begin{align} & =5\times 1000 \\ & =5000\text{cm} \\ \end{align}\]

Q2. The length and breadth of a rectangular box is in the ratio \[4:5.\] If its perimeter \[4\text{m }40\text{cm}.\] Find its dimensions.

Ans:

Given: $\frac{l}{b}=\frac{4}{5}$

Perimeter $=4\,\text{m }40\,\text{cm}$

We need to find the length and breadth of the rectangular box.

Let length of the box $=4x$

Breadth of the box $=5x$

So, we know that

\[\begin{align} & \text{P}=2(l+b)=4\text{m }40\text{cm} \\ & =2(4x+5x)=4\text{m }40\text{cm} \\ & =2(9x)=4.4\text{m} \\ & =18x=4.4\text{m} \\ & \Rightarrow x=\frac{4.4}{18} \\ & =0.244\text{m} \\ \end{align}\]

Therefore, the dimensions of the rectangular box will be

\[\begin{align} & l=4x \\ & =4\times 0.244 \\ & =0.976\text{ cm} \\ & =97.6\text{ m} \\ & b=5x \\ & =5\times 0.244 \\ & =1.22\text{ cm} \\ & =122\text{ m} \\ \end{align}\]

Q3. How many square tiles each of side \[0.25\,\text{m}\] will be required to pane the floor of a room which is \[4\,\text{m}\]broad and \[3\,\text{m}\] long?

Ans: 

Given: Length of room, $l=4\,\text{m}$

Breadth of the room, $b=3\,\text{m}$

Area of the room will be

\[\begin{align} & =4\times 3\,\text{square meters} \\ & =12\,\text{square meters} \\ \end{align}\]

Side of square tile =0.25 m

\[\begin{align} & =\frac{1}{4}\,\text{m} \\ \end{align}\]

Area of each square tile will be

\[\begin{align} & =\left( \frac{1}{4}\times \frac{1}{4} \right)\,\text{sq}\text{. meters} \\ & =\frac{1}{16}\,\text{sq}\text{. meters} \\ \end{align}\]

Therefore, number of tiles needed to pane the floor of the room will be

\[\begin{align} & =12\div \frac{1}{16} \\ & =12\times 16 \\ & =192\,\,\text{tiles} \\ \end{align}\]

Q4. Calculate area( all units are in m)

Given:

We need to find the area of the figure.

We know that area of rectangle $=l\times b$

So, area of rectangle, $1={{A}_{1}}=l\times b$

\[\begin{align} & =3\times 1 \\ & =3\text{ sq}\text{. meters} \\ \end{align}\]

Area of rectangle, $2={{A}_{2}}=l\times b$

\[\begin{align} & =6\times 1 \\ & =6\text{ sq}\text{. meters} \\ \end{align}\]

Area of rectangle, $3={{A}_{3}}=l\times b$

\[\begin{align} & =3\times 1 \\ & =3\text{ sq}\text{. meters} \\ \end{align}\]

Therefore, area of the figure will be

\[\begin{align} & A={{A}_{1}}+{{A}_{2}}+{{A}_{3}} \\ & =3+6+3 \\ & =12\text{ sq}\text{. meters} \\ \end{align}\]

Long Answer Questions. 4 Marks                              

Q1. Find the perimeter of a regular heptagon having each side equal to \[5.5\text{ cm}\]

Ans:

Given: A regular heptagon with side \[5.5\text{ cm}\]

We need to find the perimeter of the heptagon.

We know that perimeter is the length of outer sides of a figure.

Regular heptagon means a figure of $7$ sides and all sides are equal.

Therefore, the perimeter of the heptagon will be

\[\begin{align} & =7\times \text{side} \\ & =7\times 5.5 \\ & =38.5\text{ cm} \\ \end{align}\]

Q2. Find the perimeter of the below figure and the figure is symmetrical about its horizontal axis.

Ans:

Given: the figure which is symmetrical about its horizontal axis.

We need to find the perimeter of the given figure.

We know that the perimeter is the total length of all sides of the given figure.

Therefore, perimeter of the given figure will be

\[\begin{align} & =8+16+4+12+12+4+16 \\ & =8+20+21+20 \\ & =70\text{ cm} \\ \end{align}\]

Q3. Find the area of the following figure with each square of area \[1\text{ c}{{\text{m}}^{\text{2}}}\].

Ans:

Given: The figure having each square of area \[1\text{ c}{{\text{m}}^{\text{2}}}\]

We need to find the area of the figure.

Total number of squares the figure have $=16$

Area of one square \[=1\text{ c}{{\text{m}}^{\text{2}}}\]

Therefore, the area of the figure will be

\[\begin{align} & =16\times 1\text{ c}{{\text{m}}^{\text{2}}} \\ & =16\text{ c}{{\text{m}}^{\text{2}}} \\ \end{align}\]

Q4. Find area of rectangle whose length and breadth are \[35\text{ cm}\] and \[15\text{ cm}\]. Find the perimeter of the rectangle.

Ans:

Given: length of rectangle, \[l=35\text{ cm}\]

Breadth of rectangle, $b=15\text{ cm}$

We need to find the area and perimeter of the rectangle.

We know that area of rectangle $=l\times b$

Therefore, area will be

\[\begin{align} & =35\times 15 \\ & =525\text{ c}{{\text{m}}^{2}} \\ \end{align}\]

Also, we know that perimeter of a rectangle $=2(l+b)$

Therefore, perimeter will be

\[\begin{align} & =2(35+15) \\ & =2\times 50 \\ & =100\text{ cm} \\ \end{align}\]

Long Answer Questions.                                              5 Marks                              

Q1. The cost of cultivating a square field is \[\text{Rs}.130\text{ per meter}\] is \[\text{Rs}\text{. }8,450\] . Find the length of each side of the field.

Ans: 

Given: Cost of cultivating a square field $=\text{Rs}.130\text{ per meter}$

Total cost of cultivating the square field $=\text{Rs}\text{. }8,450$

We need to find the length of the side of square field.

Area of square field $=\frac{\text{Total cost of cultivating}}{\text{Per meter cost of cultivating}}$

Therefore, area will be

\[\begin{align} & =\frac{8450}{130} \\ & =65\text{ }{{\text{m}}^{2}} \\ \end{align}\]

Now, we know that area of a square $=\text{side}\times \text{side}$

Therefore, the length of the each side of square field will be

\[\begin{align} & 65=\text{side}\times \text{side} \\ & {{\text{s}}^{\text{2}}}=65 \\ & \text{s}=\sqrt{65} \\ & \text{s}=8.06\text{ cm} \\ \end{align}\]

Mensuration Class 6 Important Questions Free PDF Download

Class 6 maths chapter 10 important questions set can be downloaded in PDF format entirely free of cost. However, you can even study the same from Vedantu's website. All the problems are worked out as per the latest CBSE rules and written in a step by step manner, making it feasible for everyone to understand.

Mensuration Questions for Class 6 Overview

Important Questions for Class 6 Maths Chapter 10: Questions Carrying 1 Mark

In the first section of class 6 mensuration questions PDF, you will see eight fill in the blank questions of 1 mark each. It is one of the easiest portions, and you must be thorough with the formulas and concepts to precisely answer these questions.

Important Questions for Class 6 Maths Chapter 10: Questions Carrying 2 Marks

This segment in the PDF of important questions for class 6 maths mensuration contains 5 questions of 2 marks each. Here, you are supposed to solve the problems and find out the perimeter, area, cost of fencing a field, etc. of different shapes like square, rectangle and circle. Hence, knowing the right formula is immensely important to get correct answers. 

You can refer to the solutions provided in this PDF and recapitulate knowledge of the mensuration chapter.

Important Questions for Class 6 Maths Chapter 10: Questions Carrying 3 Marks

Here, in questions on mensuration for class 6 PDF, the questions are complicated than the previous ones. For instance, a rectangular field's length and width are provided, and it is said that a watchman has walked 5 rounds. You are asked to find out the total distance he covered.

Similar type of questions are there in the important question material, and you must go through each sum to be thorough with the solution procedures.

Important Questions for Class 6 Maths Chapter 10: Questions Carrying 4 Marks

This section in CBSE class 6 maths mensuration questions consist of four questions, out of which three of them are diagram based. Figures with some of the dimensions are given, and you are required to find out the perimeter, area, etc. of the same.

Important Questions for Class 6 Maths Chapter 10: Questions Carrying 5 Marks

The final section contains only three questions. First one is related to finding the length of one side of a field, the second one is about evaluating the perimeter of a given rectangle, and the last one requires you to calculate the breadth of a room.

All the necessary dimensions are provided; all you need to do is take the right approach and put the right formula to solve the sums.

So, make sure to download Vedantu's important questions for class 6 maths chapter 10 and revise the techniques to solve problems more quickly.

What are the Benefits of Important Questions from Vedantu for Class 6 Maths Chapter 10 - Mensuration

Explore the advantages of Vedantu's Important Questions for Class 6 Maths Chapter 10 - "Mensuration." Tailored to the chapter's essence, these benefits include focusing on key topics for efficient study, alleviating exam anxiety, reinforcing fundamental concepts, imparting effective time management skills, enabling self-assessment and progress tracking, fostering a strategic approach for higher scores, covering a broad range of topics for comprehensive understanding, and providing crucial support for exam preparation, instilling confidence in class 7 students with simplicity and clarity.

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